Question

The frequency of light emitted for the transition $n=4ton=2$ of the $H{e}^{+}$ is equal to the transition in $H$ atom corresponding to which of the following ?
(correct answer +1, wrong answer -0.25)

• A. $n=2ton=1$
• B. $n=3ton=2$
• C. $n=4ton=3$
• D. $n=3ton=1$

Answer 1

The correct option is A $n=2ton=1$

For $H{e}^{+}$
$\overline{\nu }=\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{2^2}-\frac{1}{4^2})$
For $H$
$\overline{\nu }=\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
For same frequency,
$z^2(\frac{1}{2^2}-\frac{1}{4^2})=(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Since, $z=2$
$(\frac{1}{n_1^2}-\frac{1}{n_2^2})=(\frac{1}{1^2}-\frac{1}{2^2})$
$n_1=1and{n}_2=2$